Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 31

$$\eqalign{ & {z_{xx}}\left( {x,y} \right) = \frac{1}{x},\,\,\,\,{z_{yy}}\left( {x,y} \right) = - \frac{x}{{{y^2}}} \cr & {z_{xy}}\left( {x,y} \right) = {z_{yx}}\left( {x,y} \right) = \frac{1}{y} \cr} $$

$$\eqalign{ & z = x\ln \left| {xy} \right| \cr & {\text{Find }}{z_x}\left( {x,y} \right){\text{ and }}{z_y}\left( {x,y} \right) \cr & {z_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {x\ln \left| {xy} \right|} \right] \cr & {\text{treat }}y{\text{ as a constant and }}x{\text{ as a variable}}{\text{. then by the power rule}} \cr & {z_x}\left( {x,y} \right) = x\frac{\partial }{{\partial x}}\left[ {\ln \left| {xy} \right|} \right] + \ln \left| {xy} \right|\frac{\partial }{{\partial x}}\left[ x \right] \cr & {z_x}\left( {x,y} \right) = x\left( {\frac{1}{{xy}}} \right)\left( y \right) + \ln \left| {xy} \right|\left( 1 \right) \cr & {z_x}\left( {x,y} \right) = 1 + \ln \left| {xy} \right| \cr & \cr & {z_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {x\ln \left| {xy} \right|} \right] \cr & {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then}} \cr & {z_y}\left( {x,y} \right) = x\frac{\partial }{{\partial y}}\left[ {\ln \left| {xy} \right|} \right] \cr & {z_y}\left( {x,y} \right) = x\left( {\frac{1}{{xy}}} \right)\frac{\partial }{{\partial y}}\left[ {\left| {xy} \right|} \right] \cr & {z_y}\left( {x,y} \right) = x\left( {\frac{1}{{xy}}} \right)\left( x \right) \cr & {z_y}\left( {x,y} \right) = \frac{x}{y} \cr & \cr & {\text{find the second - order partial derivatives}} \cr & {z_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {1 + \ln \left| {xy} \right|} \right] \cr & {z_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ 1 \right] + \frac{\partial }{{\partial y}}\left[ {\ln \left| {xy} \right|} \right] \cr & {z_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ 1 \right] + \frac{1}{{xy}}\frac{\partial }{{\partial y}}\left[ {xy} \right] \cr & {z_{xy}}\left( {x,y} \right) = 0 + \frac{1}{{xy}}\left( x \right) \cr & {z_{xy}}\left( {x,y} \right) = \frac{1}{y} \cr & \cr & {z_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{x}{y}} \right] \cr & {z_{yx}}\left( {x,y} \right) = \frac{1}{y}\frac{\partial }{{\partial x}}\left[ x \right] \cr & {z_{yx}}\left( {x,y} \right) = \frac{1}{y}\left( 1 \right) \cr & {z_{yx}}\left( {x,y} \right) = \frac{1}{y} \cr & \cr & {z_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {1 + \ln \left| {xy} \right|} \right] \cr & {z_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ 1 \right] + \frac{\partial }{{\partial x}}\left[ {\ln \left| {xy} \right|} \right] \cr & {z_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ 1 \right] + \left( {\frac{1}{{xy}}} \right)\frac{\partial }{{\partial x}}\left[ {xy} \right] \cr & {z_{xx}}\left( {x,y} \right) = 0 + \left( {\frac{1}{{xy}}} \right)\left( y \right) \cr & {z_{xx}}\left( {x,y} \right) = \frac{1}{x} \cr & {\text{and}} \cr & {z_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{x}{y}} \right] \cr & {z_{yy}}\left( {x,y} \right) = - \frac{x}{{{y^2}}} \cr} $$